Exactly! And with wind-shift surely... http://t3.gstatic.com/images?q=tbn:ANd9GcT64bvZTta0cZAwnBcrlt7VjloO4sbwGF6F0tHTj6JEJMWIcAHNIg + http://t1.gstatic.com/images?q=tbn:ANd9GcRevJMOgBclkCzIpoykCNV1Ku05MY3g65QaZ227z-RUt49h63PDpw > http://i.dailymail.co.uk/i/pix/2012/01/17/article-2087814-0F8035B200000578-692_634x396.jpg OR A spherical teardrop of mass m, radius r, vertical velocity v (positive downward), and mass density ρd ≈ 1 g cm−3 satisfies m = 4 3 πρdr3 (1) dm dt = πρmr2|v| (2) m dv dt = mg − C 2 πρar2v|v| − v dm dt (3) where the terms on the right-hand side of equation (3) give the weight, air drag, and mist drag acting on the growing drop. When the air drag is written as above, the drag coefficient C = C® is a function only of the Reynolds number [6, 7] R = 2r|v|/ν, where ν is the kinematic viscosity of air. Figure 1 shows C® measured for falling liquid drops (data points, [8]), together with the small-R theoretical Stokes result C = 24/R (broken line, [6]) and a new empirical result (full line), C = 12R−1/2 (4) valid over the intermediate range 10 < R < 1000. Such simple algebraic results allow us to avoid complicated numerical results for C® obtained from the nonlinear Navier?Stokes equations [7]. Inserting equation (4) into equation (3) and setting dm/dt = dv/dt = 0 yields the mist-free terminal or ?settling? speed of drops of radius r with 10 < R < 1000: v0 = σr (5)